Diagonalize [3, 1; 0, 2]

Step by step

The question

Question: diagonalize A.

Diagonalizing means writing A as P times D times P inverse.

The columns of P are eigenvectors.

D is diagonal: the eigenvalues down the diagonal, 0s elsewhere.

1 Find the eigenvalues

As in the eigenvalue lesson: take λ off the diagonal of A.

Its determinant, down minus up, must be 0.

So (3 − λ)(2 − λ) − 0 × 1 = 0.

Multiply out the brackets.

Like terms together, the highest power first: λ² − 5λ + 6 = 0.

It factors, as in the factoring lesson.

So λ = 2 or λ = 3.

2 Eigenvector for λ = 2

We take 2 off the diagonal: 3 − 2 = 1 and 2 − 2 = 0.

A row of 0s: the letter y is free.

That is why there are eigenvectors.

Row 1 as an equation: x + y = 0.

We want x alone.

To remove + y, we take away y on both sides.

They cancel on the left.

So x = −y.

We pick y = 1.

So an eigenvector for λ = 2 is (−1, 1).

3 Eigenvector for λ = 3

We take 3 off the diagonal: 3 − 3 = 0 and 2 − 3 = −1.

A row of 0s: the letter x is free.

That is why there are eigenvectors.

Row 1 as an equation: y = 0.

So y = 0.

We pick x = 1.

So an eigenvector for λ = 3 is (1, 0).

4 Build P and D

P: the eigenvector for λ = 2 is column 1, the one for λ = 3 is column 2.

D: 2 first, then 3, in the same order.

Careful!

The order matters.

The first λ in D goes with the first column of P.

So A = P D P⁻¹, with this P and D.

5 Check

Let's check without the inverse: A = P D P⁻¹ means A P = P D.

We work out both sides.

First A times P.

Now P times D.

Both give rows −2, 3 and 2, 0.

So A P = P D, and the answer is right!

Let's look back.

First: find the eigenvalues.

Then: eigenvector for λ = 2.

Then: eigenvector for λ = 3.

Then: build P and D.

Last: check.